Adding Ten Tiles I

The Problem

Jim has ten tiles with one of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 on each of them. He plays around and discovers that he can make some addition sums like the one below. What’s more, the four numbers on the tiles in the answer to the sum (the numbers in the bottom row), add up to 9.

How many sums like that can he make?

Teaching sequence

1. Start with the class in their groups and either pose the problem to the whole class or give each group a copy of the problem and tell them to begin. Check that they all understand that there are ten different numbers (0 to 9 inclusive) that have to be used in Jim’s problem.
2. Allow the students to guess at first and record their answers on the board as they find them. For groups who seem to be stuck, ask
   Where do you think the number 1 goes? Why?
   What four numbers add up to 9? So what could just the answer to Jim’s sum be?
3. For groups that have managed to guess a few answers ask
   Have you checked the addition for your answer?
   How did you get the answer?
4. When two different addition answers with the same digits in the sum have been found ask
   What are the similarities between these answers?
   Can you get another addition sum using the same numbers? How many?
5. When a number of answers have been found, get the class’s attention. Ask
   Have we got all of the possible answers?
   How will we know when we have all of the answers?
6. Ask some of the groups to explain what they have done. Get all groups to write up the problem as far as they were able to go, giving reasons for what they have done.

Extension to the problem

Show that you have found all possible answers to Jim’s problem.

Solution

To start this problem off we would be inclined to guess and check and hope that something turns up. The full set of answers is:

How can we get all of these answers and be sure that there are no others? We can either do this algebraically or by using a systematic approach. We delay the algebraic approach until Adding Ten Tiles II (though there is no reason why you can’t look up the solution there) and use the systematic approach here.

Step 1: The first thing to notice is the value of the left-most digit in the answer. It couldn’t be 0 as then the answer wouldn’t be a four-digit answer. In that case we’d be looking at the problem Nine Tiles.

Now to get that digit we have to add two other digits plus a possible carry-over from the tens column. The biggest two digits we could have are 8 and 9. The carry-over cannot be bigger than 2. The sum of 8 and 9 and 2 is 19, which is less than 20. So the carry-over from the hundreds column is 1. This is the value of the left-most number in the answer.

Step 2: Now we have to get an answer whose digits add up to 9. But one of these digits is 1. So we have to find three different digits whose sum is 8. Doing this systematically we get the following two possibilities: 6, 2, 0; 5, 3, 0. (Remember that 5, 2, 1 is not possible, as we have already used up the 1.)

So the answer to the Jim’s sum might be 1026, 1062, 1206, 1260, 1602, 1620 or 1035, 1053, 1305, 1350, 1503, 1530.

Step 3: Getting the answers. Let’s try 1026 first. To get a 6 the only combination that will work is 7 + 9. (0 + 6, 1 + 5, 2 + 4, 3 + 3, 8 + 8 won’t work for various reasons – either we have already used up one of the digits in the sum or the same digit is needed twice.) How do we get a 2? Well it’s really 1 because we have a carry-over of 1 already. The only way that is legal is 3 + 8. Then we have to find two digits that with 1 give 10. So we want two digits whose sum is 9. But the only digits we haven’t used yet are 4 and 5 and they fit the bill nicely. So we get the answer (i) from above or various combinations like

and so on. But to make things simpler we will agree that they are all equivalent because once we have one we can then get all the others just by moving the numbers around.

Working with the other four-digit answers we get the eight answers we listed above, give or take the variations we’ve already seen in the columns for 1026.

Step 4: So there are 8 answers to Jim’s problem (or 64 if you count all the different equivalent answers).

Solution to the extension

Note that in doing the solution to Jim’s problem carefully, we have answered the Extension problem too.